Question

Show that two noncommuting operators cannot have a complete set of common eigenfunctions. Hint: Show that if $\hat{\mathbf{P}}$and $\hat{\mathbf{Q}}$have a complete set of common eigenfunctions, then $\left[\hat{P}.\hat{Q}\right]\mathbf{f}\mathbf{=}\mathbf{0}$for any function in Hilbert space.

Short answer

A whole set of common eigenfunctions cannot be shared by two noncommuting operators.

Step-by-step solution

Concept used

The same complete set of common eigenfunctions:

$\hat{\mathbf{P}}{\mathbf{f}}_{\mathbf{n}}\mathbf{=}{\mathbf{\lambda }}_{\mathbf{n}}{\mathbf{f}}_{\mathbf{n}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{and}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\hat{\mathbf{Q}}{\mathbf{f}}_{\mathbf{n}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{n}}{\mathbf{f}}_{\mathbf{n}}$

Calculation

Assuming the operators $\hat{P}$ and $\hat{Q}$have the same complete set of common eigenfunctions, that is:

$\hat{\mathrm{P}}{\mathrm{f}}_{\mathrm{n}}={\mathrm{\lambda }}_{\mathrm{n}}{\mathrm{f}}_{\mathrm{n}}\mathrm{and}\hat{\mathrm{Q}}{\mathrm{f}}_{\mathrm{n}}={\mathrm{\mu }}_{\mathrm{n}}{\mathrm{f}}_{\mathrm{n}}$

And suppose the set $f_n$is complete, so that any function in Hilbert space $f\left\{x\right)$can be expressed as a linear combination, that is:

$f=\sum c_n{f}_n$

Solving for the above function,

$$\begin{gathered} \left[\hat{P},\hat{Q}\right]f=\left(\hat{P}\hat{Q}-\hat{Q}\hat{P}\right)\sum c_n{f}_n \\ =\hat{P}\left(\sum c_n{\mu }_n{f}_n\right)-\hat{Q}\left(\sum c_n{\lambda }_n{f}_n\right) \\ =\sum c_n{\mu }_n{\lambda }_n{f}_n-\sum c_n{\lambda }_n{\mu }_n{f}_n \\ =0 \end{gathered}$$

If two operators have the same set of eigenfunctions, the commutator will be zero.